Then, to open : Pxx/i f —— 6 Appendix VI. and the opening stress/ == 1093 And to close : P = closing stress being / x bti __ bh But these must balance : bh ~ bJi and x = - so P cannot be placed further from the centre than -J ^, or the line of pressure must be kept within the middle third of the voussoir depth h. The Masonry Arch, Fig. 966, is drawn in half only, the loads being symmetrically disposed; and there are five voussoirs, v v &c., to each half. The loads w <w &c, are found by erecting perpendiculars a b c d ef, each load being the resultant of the weight of wall between a pair of verticals combined with the weight of the voussoir, the latter being, of course, propor- tionately small. Set these loads out in the force diagram as 01, 12, 23, &c. By symmetry the force O o must be horizontal. Take any pole Ox and draw the link polygon g h, then find r s the resultant of Oo and 04 the pressures at skewback and key respectively. Take points / and k and find a new polygon passing through them, such that the link 0 o still remains horizontal and meeting r s in /. Join / k and draw the line 4 02 || / k, thus giving the new pole 02. Complete the polygon k /, which will indicate the line of pressure or resistance between the voussoirs, the magnitudes of the forces being found from the force diagram. The line kl must lie within the middle third of the joint, as shewn dotted, and if this does not occur everywhere a readjustment must be made. The extreme possible conditions are shewn at E and F, where the line m n is better than f q^ because it causes a lower value for the pressures. The greatest pressure occurs at the skewback or support, and the intensity must never be more than the material will allow. If the line of resistance passes through the centre the intensity is considered uniform, but if the line touches *the dotted arc the maximum intensity will be twice that of the mean.